Question: The following equation is true for all real values of $y$ for which the expression on the left is defined, and $D$ is a polynomial expression. $\dfrac{20y^2-80}{D} \div \dfrac{4y^2-8y}{y^3+9y^2}=1$ What is $D$ ? $D=$
Explanation: The left side of the equation is a quotient of two rational expressions, while the right side is simply $1$. This means that the numerator and the denominator of the resulting quotient on the left side should cancel out completely. In order to solve for $D$, let's divide the expressions and simplify as much as we can. We start with factoring the numerators and the denominators of the two expressions. [Why are we doing this?] The numerator, $20y^2-80$, of the dividend can be factored as $20(y+2)(y-2)$ by factoring out $20$ and using the difference of squares pattern. The numerator, $4y^2-8y$, of the divisor can be factored as $4y(y-2)$ by factoring out a $4y$. The denominator, $y^3+9y^2$, of the divisor can be factored as $y^2(y+9)$ by factoring out $y^2$. Now the quotient looks as follows: $\dfrac{20(y+2)(y-2)}{D} \div \dfrac{4y(y-2)}{y^2(y+9)}$ To find the quotient of two rational expressions, we flip the divisor, multiply across, then simplify: [What's that?] $\begin{aligned} &\phantom{=} \dfrac{20(y+2)(y-2)}{D} \div \dfrac{4y(y-2)}{y^2(y+9)} \\\\\\ &= \dfrac{20(y+2)(y-2)}{D} \cdot \dfrac{y^2(y+9)}{4y(y-2)} &\text{Flip the divisor} \\\\\\ &= \dfrac{20(y+2)(y-2) \cdot y^2(y+9)}{D \cdot 4y(y-2)} &\text{Multiply across.}\\\\\\ &= \dfrac{5\cdot {\cancel{4}}(y+2){\cancel{(y-2)}} {\cancel{y}} \cdot y(y+9)}{D {\cancel{4}}{\cancel{y}}{\cancel{(y-2)}}} &\text{Cancel out common factors.}\\\\\\\\ &=\dfrac{5y(y+2)(y+9)}{D} \end{aligned}$ After this simplification, our equation now looks like: $\dfrac{5y(y+2)(y+9)}{D}=1$ We can conclude that the numerator and the denominator of the expression on the left must be equal. [Why?] In other words, $D=5y(y+2)(y+9)$, which is equivalent to $5y^3+55y^2+90y$.